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I am trying to show two equivalent metrics $p$ and $d$ on a set $X$ have the same convergent sequences. $p$ and $d$ are such that $kd(x,y) \leq p(x,y) \leq td(x,y)$ for every $x, y \in X$, $k$ and $t$ are positive constants.

Here's what I am doing -

As $p$ and $d$ are equivalent metrics they generate the same open sets.

Let $A$ be an open set generated by both $p$ and $d$.

As $A$ is open $X\backslash A$ is closed.

As $X\backslash A$ is closed, we can take a convergent sequence $(x_n) \in X\backslash A$ for all $n$, and it will converge to $x \in X\backslash A$.

Im not sure what to do now...can you just say it will converge to the same $x$ regardless of the metric being $p$ or $d$? I don't think I can. Should I be bringing open balls into it? Is there a need to use the positive constants $k$ and $t$?

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up vote 4 down vote accepted

If you’ve really already proved that metrics $p$ and $d$ related in that way generate the same open sets, you’re practically done, but you’re trying to make it much too complicated. Suppose that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ with respect to $d$; you want to show that it converges to $x$ with respect to $p$ as well. Let $U$ be an open nbhd of $x$. Then since $\langle x_n:n\in\Bbb N\rangle\underset{d}\longrightarrow x$, there is an $m\in\Bbb N$ such that $x_n\in U$ for all $n\ge m$. But that’s also exactly what it means for $\langle x_n:n\in\Bbb N\rangle$ to converge to $x$ with respect to $p$, so $\langle x_n:n\in\Bbb N\rangle\underset{p}\longrightarrow x$.

It’s in the proof that $d$ and $p$ generate the same topology that you would use the constants $k$ and $t$. But it’s not necessary to prove first that $d$ and $p$ generate the same topology: you can prove this result directly.

Suppose that $\langle x_n:n\in\Bbb N\rangle\underset{d}\longrightarrow x$. Then for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $d(x_n,x)<\epsilon$ for each $n\ge m_\epsilon$. This immediately implies that $p(x_n,x)<t\epsilon$ for each $n\ge m_\epsilon$. Thus, for each $n\ge m_{\epsilon/t}$ we have $p(x_n,x)<t\cdot\frac{\epsilon}t=\epsilon$, and it follows that $\langle x_n:n\in\Bbb N\rangle\underset{p}\longrightarrow x$. The opposite implication is proved similarly, using the fact that $d(x,y)\le\frac1kp(x,y)$ for all $x,y\in X$.

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I am not sure how your open paragraph proves they have the same convergent sequences. It seems like you said "The sequence $(x_n)$ converges to $x$ with respect to $d$. It also converges to $x$ with respect to $p$". It seems like more of a statement than a proof, as if it is just stated that it converges to $x$ with respect to $p$ as opposed to proving it. – sonicboom Nov 6 '12 at 20:07
@sonicboom: That’s because it really is that trivial once you know that $d$ and $p$ generate the same open sets. To show that if the sequence converges to $x$ with respect to $d$, it also does so w.r.t. $p$, you start with a $p$-open nbhd $U$ of $x$; it’s also $d$-open, so a tail of the sequence is inside it. This is true for every $p$-open nbhd of $x$, so the sequence does indeed converge to $x$ w.r.t. $p$. As I said, the real work in this approach is in proving that $d$ and $p$ generate the same open sets. – Brian M. Scott Nov 6 '12 at 20:11

Note that your description $$kd(x,y)\leq p(x,y)\leq td(x,y)\tag{1}$$ means that $p$ and $d$ are strongly equivalent metrics. For such metrics, it's fairly easy to show that we have the same convergent sequences.

Let $(x_n)$ be any sequence of points of $X$.

On the one hand, suppose that $p(x_n,x)\to0$ for some $x\in X$. Take any $\epsilon>0$. To show that $d(x_n,0)\to 0$, we must show that there exists some $N_1$ such that $d(x_n,x)<\epsilon$ for all $n\geq N_1$. Since $p(x_n,x)\to 0$, then there exists $N_1$ such that $p(x_n,x)<k\epsilon$ for all $n\geq N_1$, and so by $(1)$, $$d(x_n,x)=\frac1k\cdot kd(x_n,x)\leq\frac1k\cdot p(x_n,n)<\frac1k\cdot k\epsilon=\epsilon$$ for all $n\geq N_1$.

On the other hand, suppose $d(x_n,x)\to 0$ for some $x\in X$. Taking $\epsilon>0$ and finding $N_2$ such that $d(x_n,x)<\frac{\epsilon}t$ for all $n\geq N_2$, we again use $(1)$ to see that $p(x_n,x)<\epsilon$ for $n\geq N_2$.

Thus, the metrics share the same convergent sequences.

It isn't actually necessary to use the constants $k,t$--in fact, it's even okay if there are no positive constants $k,t$ satisfying $(1)$ for all $x,y\in X$.

We say that metrics are equivalent if they produce the same open sets. Put rigorously, we mean the following:

(i) $\forall x,y,z\in X$ $\forall R>0$, if $d(x,y)<R$, then there exists $r>0$ such that $d(x,z)<R$ whenever $p(y,z)<r$. (Every point in an open $d$-ball lies in an open $p$-ball contained in the $d$-ball.)

(ii) [Same thing as (i), but swapping $p$ and $d$ in each instance.]

Strongly equivalent metrics are equivalent, but the converse needn't hold.

Suppose that $p$ and $d$ are equivalent, and let $(x_n)$ is a sequence of points of $X$.

On the one hand, suppose $p(x,x_n)\to 0$, and take $\epsilon>0$. Since $p$ and $d$ are equivalent, then there is some $r>0$ such that $d(x,x_n)<\epsilon$ whenever $p(x,x_n)<r$. Since $p(x,x_n)\to 0$, there is some $N$ such that $p(x,x_n)<r$ for all $n\geq N$, so $d(x,x_n)<\epsilon$ for all $n\geq N$. Thus, $d(x,x_n)\to 0$.

A similar approach will let you show that $d(x,x_n)\to0$ implies $p(x,x_n)\to 0$.

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Ah, I thought you look for examples:

  1. The easiest if you take any metric $d$, and its double $d'(x,y):=2d(x,y)$.
  2. Other classic examples are the metrics on $\Bbb R^n$ given by the $||.||_p$ norms ($p\ge 1$): $$||(a_1,..,a_n)||_p:= \big(|a_1|^p+..+|a_n|^p \big)^{1/p}$$ Draw the unit circles for $n=2$ and $p=1,2,\infty$. Where $||(a_1,..,a_n)||_\infty = \displaystyle\max_i |a_i|$. These all generate the same metric (metric from norm by $d(x,y):=||x-y||$).

Anyway, for the convergent sequnces, use this $$x_n\to x \iff d(x_n,x)\to 0$$

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Keep in mind that in a metric space with metric $d$, a sequence $(x_n)_{n=1}^{\infty}$ converges to $x \Leftrightarrow$ if $U$ is an open set (with respect to the topology generated by $d$) containing $x$ then $\exists N \in \mathbb{Z}^+$ such that $x_n \in U$ for all $n \geq N$.

So, suppose $(x_n)_{n=1}^{\infty}$ converges to $x$ with respect to the metric $d$. Let $U$ be an open set (in the topology induced by $p$) containing $x$. $U$ is open in the topology generated by $d$. (You already know that equivalent metrics generate the same topology.) Thus $\exists N \in \mathbb{Z}^+$ such that $x_n \in U$ for all $n \geq N$. This shows that $(x_n)_{n=1}^{\infty}$ converges to $x$ with respect to the metric $p$.

You can also go in the other direction, to show that convergence to $x$ with respect to $p$ implies convergence with respect to $d$.

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Hm. I guess the correct answer would depend on how exactly you define convergence. The way I'm used to is like this: metric -> topology -> convergence. That is, given a metric, you define a topology (i.e. you define which sets will be called open and which will not). When you have a topology, you define convergence.

If you follow this approach, then you were already done at the phrase "they generate the same open sets". If two metrics give you the same open sets (i.e. the same topology), then they will also give you the same convergent sequences, because convergence is defined in terms of open sets and nothing else.

However, if your approach is different and you define convergence using the metric directly, then proving your statement will become a technical task of verifying the definitions.

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In a metric space a sequence $(x_n)_{n\in N}$ converges to $x$ iff for every nbhd $U$ of $x,$ the set $\{n\in N :x_n\not \in U\}$ is finite. If two metrics generate the same topology, the set of nbhds of $x$ is the same for both metrics, and hence the metrics generate the same convergent sequences.

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